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Computers and Mathematics with Applications
journal homepage:www.elsevier.com/locate/camwaThe solution of multipoint boundary value problems by the Optimal
Homotopy Asymptotic Method
Javed Ali
a, S. Islam
a,b,∗, Sirajul Islam
c, Gul Zaman
daIslamia College Peshawar (Chartered University), NWFP, Pakistan bDepartment of Mathematics, CIIT, H-8/, Islamabad, Pakistan cUET, Peshawar, NWFP, Pakistan
dCAMP, NUST, Rawalpindi, Pakistan
a r t i c l e i n f o
Article history:
Received 9 June 2009
Received in revised form 24 November 2009
Accepted 1 December 2009
Keywords:
Multipoint boundary value problems Optimal Homotopy Asymptotic Method
a b s t r a c t
In this work, we solve multipoint boundary value problems using the Optimal Homotopy Asymptotic Method (OHAM).The proposed method is tested upon several problems from the literature and the results are compared with the available exact solution. This method provides easy tools to control the convergence region of approximating solution series where ever necessary.
© 2009 Elsevier Ltd. All rights reserved.
1. Introduction
Real world physical problems are generally described by differential equations. These equations are often handled by the most common methods; Adomian decomposition method (ADM) [1,2], differential transform method (DTM) [3], variational iteration method (VIM) [4], successive iteration [5], splines [6], homotopy perturbation method (HPM) [7–13], homotopy analysis method (HAM) [14] etc. Perturbation methods are based on small or large parameters and hence cannot deal with strong nonlinearity in the problems. The nonperturbation methods like ADM and DTM can deal with strongly nonlinear problems but the convergence region of their series solution is generally small. The HPM accompanying homotopy and the perturbation technique overcomes the restrictions of small or large parameters in the problems. It deals with nonlinear problems effectively.
Recently Vasile Marinca et al. [15–18] introduced OHAM for the approximate solution of nonlinear problems of thin film flow of a fourth-grade fluid down a vertical cylinder. In their work they have used this method to understand the behavior of nonlinear mechanical vibration of an electrical machine. They also used the same method for the solution of nonlinear equations arising in the steady state flow of a fourth-grade fluid past a porous plate and for the solution of a nonlinear equation arising in heat transfer. Furthermore, they have also shown that HPM and HAM are the special cases of OHAM. This method is straight forward, reliable and it does not need to look for h curves like HAM. This method provides a convenient way to control the convergence of the series solution and allows the adjustment of convergence region where ever it is needed. The OHAM solution generally agrees with the exact solution at large domains as compared to HPM and HAM solutions. We use this method to find the approximate analytic solution of multipoint BVPs. These BVPs appear in modeling the variations of a guy wire of uniform cross-section and composed of N parts of different densities. The theory of elastic stability is handled by multipoint problems for several cases in [19]. Some theoretical works regarding these problems
∗Corresponding author at: Department of Mathematics, CIIT, H-8/, Islamabad, Pakistan. Tel.: +92 333 9844540; fax: +92 51 4442805.
E-mail address:[email protected](S. Islam).
0898-1221/$ – see front matter©2009 Elsevier Ltd. All rights reserved. doi:10.1016/j.camwa.2009.12.002
are presented in [20–22]. Substantial theoretical work has been done for these problems but very little had been done for the numerical solution. For some numerical methods we refer the reader to see [23–25]. The results of OHAM are compared with those of exact solution and portrayed graphically.
The structure of this paper is organized as follows. Section2is devoted to the analysis of the proposed method. Some numerical examples are presented in Section3. In Section4, we concluded by discussing results of the numerical simulation by using Mathematica.
2. Analysis of the method
Consider the following differential equation:
L
(
u(
x)) +
g(
x) +
N(
u(
x)) =
0,
B u,
du dx=
0,
(2.1)where L is a linear operator, x denotes independent variable, u
(
x)
is an unknown function, g(
x)
is a known function, N is a nonlinear operator and B is a boundary operator.According to OHAM we construct a homotopy h
(v(
x,
p),
p) :
R×
[0,
1]→
R which satisfies(
1−
p)[
L(v(
x,
p)) +
g(
x)] =
H(
p)[
L(v(
x,
p)) +
g(
x) +
N(v(
x,
p))],
Bv(
x,
p),
∂v(
x,
p)
∂
x=
0 (2.2)where x
∈
R and p∈
[0,
1] is an embedding parameter, H(
p)
is a nonzero auxiliary function for p6=
0, H(
0) =
0 andv(
x,
p)
is an unknown function. Obviously, when p
=
0 and p=
1 it holds thatv(
x,
0) =
u0(
x)
andv(
x,
1) =
u(
x)
respectively. Thus, as p varies from 0 to 1, the solutionv(
x,
p)
approaches from u0(
x)
to u(
x)
, where u0(
x)
is obtained from Eq.(2.2) for p=
0 and we haveL
(
u0(
x)) +
g(
x) =
0,
B u0,
du0 dx=
0.
(2.3)Next, we choose auxiliary function H
(
p)
in the formH
(
p) =
pC1+
p2C2+ · · ·
(2.4)where C1
,
C2, . . .
are constants to be determined. H(
p)
can be expressed in many forms as reported by V. Marinca et al. [15–18].To get an approximate solution, we expand
v(
x,
p,
Ci)
in Taylor’s series about p in the following manner,v(
x,
p,
Ci) =
u0(
x) +
∞
X
k=1
uk
(
x,
C1,
C2, . . . ,
Ck)
pk.
(2.5)Substituting Eq.(2.5)into Eq. (2.2)and equating the coefficient of like powers of p, we obtain the following linear equations.
Zeroth order problem is given by Eq.(2.3)and the first order problem is given by Eq.(2.6):
L
(
u1(
x)) +
g(
x) =
C1N0(
u0(
x)),
B u1,
du1 dx=
0.
(2.6)The general governing equations for uk
(
x)
are given by: L(
uk(
x)) −
L(
uk−1(
x)) =
CkN0(
u0(
x)) +
k−1X
i=1 Ci[L(
uk−i(
x)) +
Nk−i(
u0(
x),
u1(
x), . . . ,
uk−1(
x))
],
k=
2,
3, . . . ,
B uk,
duk dx=
0 (2.7)where Nm
(
u0(
x),
u1(
x), . . . ,
um(
x))
is the coefficient of pmin the expansion of N(v(
x,
p))
about the embedding parameter p. N(v(
x,
p,
Ci)) =
N0(
u0(
x)) +
∞
X
m=1
Nm
(
u0,
u1,
u2, . . . ,
um)
pm.
(2.8)It has been observed that the convergence of the series(2.5)depends upon the auxiliary constants C1
,
C2, . . .
. If it is convergent at p=
1, one hasv(
x,
Ci) =
u0(
x) +
∞
X
k=1
uk
(
x,
C1,
C2, . . . ,
Ck).
(2.9)The result of the mth order approximations are given
˜
u(
x,
C1,
C2, . . . ,
Cm) =
u0(
x) +
mX
i=1 ui(
x,
C1,
C2, . . . ,
Ci).
(2.10)Substituting Eq.(2.10)into Eq.(2.1), it results the following residual:
R
(
x,
C1,
C2, . . . ,
Cm) =
L(˜
u(
x,
C1,
C2, . . . ,
Cm)) +
g(
x) +
N(˜
u(
x,
C1,
C2, . . . ,
Cm)).
(2.11)If R
=
0, thenu will be the exact solution. Generally it does not happen, especially in nonlinear problems.˜
In order to find the optimal values of Ci,
i=
1,
2,
3, . . .
, we first construct the functional,J
(
C1,
C2, . . . ,
Cm) =
Z
ba
R2
(
x,
C1,
C2, . . . ,
Cm)
dx (2.12)and then minimizing it, we have
∂
J∂
C1=
∂
J∂
C2= · · · =
∂
J∂
Cm=
0,
(2.13)where a and b are in the domain of the problem. With these constants known, the approximate solution (of order m) is well determined.
3. Numerical examples
3.1. Example 1
Consider the following third order linear differential equation with boundary conditions at three points [9],
y000
−
k2y0+
a=
0,
(3.1.1)y0
(
0) =
y0(
1) =
0,
y(
0.
5) =
0.
(3.1.2)Here, the physical constants are k
=
5 and a=
1.
The function y(
x)
shows the shear deformation of sandwich beams. The analytic solution of this problem is given byy
(
x) =
a k3 sinhk 2−
sinh kx+
a k2 x−
1 2+
a k3 cosh kx−
coshk 2 tanhk 2.
(3.1.3)Applying the proposed method, we have Zero Order Problem:
u0000
(
x) = −
1,
with conditions u00(
0) =
u00(
1) =
0,
y(
0.
5) =
0.
(3.1.4) It gives us,u0
(
x) = −
0.
04167+
0.
x+
0.
25 x2−
0.
16667 x3.
(3.1.5)First order problem:
u0001
(
x,
C1) = (
1+
C1)
u0000(
x) +
1+
C1−
25C1u00(
x),
with conditions u0
1
(
0) =
u0
1
(
1) =
0,
u1(
0.
5) =
0.
(3.1.6) We obtain the solution as follows:u1
(
x,
C1) = −
0.
1041667C1+
0.
x+
0.
x2+
0.
520833C1x2−
0.
520833C1x4+
0.
208333C1x5.
(3.1.7) Second order problem:u0002
(
x,
C1,
C2) = (
1+
C1)
u0001(
x,
C1) −
25 C2u00(
x) −
25C1u01(
x,
C1) +
C2+
C2u0000(
x),
with conditions, u02
(
0) =
u02(
1) =
0,
u2(
0.
5) =
0.
(3.1.8) The solution becomes:u2
(
x,
C1,
C2) = −
0.
104167C1−
0.
367684C12−
0.
104167C2+
0.
520833C1x2+
1.
82292C12x2+
0.
520833C2x2−
0.
520833C1x4−
1.
60590C12x 4−
0.
520833C 2x4+
0.
208333C1x5+
0.
208333C12x5+
0.
208333C2x5+
0.
434028C12x 6−
0.
124008C2 1x 7.
(3.1.9)Third order problem:
u0003
(
x,
C1,
C2,
C3) = (
1+
C1)
u0002(
x,
C1,
C2) −
25C1u03(
x,
C1,
C2,
C3) +
C2u01(
x,
C1)
−
25C2u01(
x,
C1) +
C3+
C3u0000(
x) −
25C3u00(
x),
Table 1
x Exact solution OHAM solution Error Error∗
0.0 −0.0121071 −0.0121071 1.298 E−10 6.653 E−5 0.1 −0.0112665 −0.0112665 −3.099 E−9 6.500 E−5 0.2 −0.00922221 −0.00922221 6.959 E−9 5.254 E−5 0.3 −0.00646687 −0.00646687 1.086 E−9 3.630 E−5 0.4 −0.00332019 −0.00332018 −1.065 E−8 1.875 E−5 0.5 0.00 3.70662 E−18 −6.155 E−17 *** 0.6 0.00332019 0.00332018 1.065 E−8 1.734 E−5 0.7 0.00646687 0.00646687 −1.086 E−9 3.405 E−5 0.8 0.00922221 0.00922221 −6.959 E−9 4.980 E−5 0.9 0.0112665 0.0112665 3.099 E−9 6.201 E−5 1.0 0.0121071 0.0121071 −1.298 E−10 6.347 E−5
Error=Exact-OHAM, Error*=Exact-Pade Approximants [25].
-2 -1 1 2 3
-1 -0.5
0.5 1
Fig. 1. The solid line (red) denotes the OHAM solution while the dotted line denotes the exact solution for the domain−2<x<3. We obtain the following solution:
u3
(
x,
C1,
C2,
C3) =
0.
104167C1−
0.
735367C12−
1.
298604C13−
0.
104167C2−
0.
62066C1C2−
0.
104167C3+
0.
x+
0.
x2+
0.
520833C1x2+
3.
645833C12x 2+
6.
418961C3 1x 2+
0.
520833C 2x2+
3.
072917C1C2x2+
0.
520833C3x2−
0.
520833C1x4−
3.
2118056C12x 4−
5.
403646C13x4−
0.
520833C2x4−
2.
647569C1C2x4−
0.
520833C3x4+
0.
208333C1x5+
0.
416667C12x 5+
0.
208333C13x5+
0.
208333C2x5+
0.
208333C1C2x5+
0.
208333c3x5+
0.
868056C12x 6+
1.
772280C13x6+
0.
850694C1C2x6−
0.
248016C12x 7−
0.
248016C3 1x 7−
0.
243056C 1C2x7−
0.
193762C13x8+
0.
043058C13x9.
(3.1.11)Using Eqs.(3.1.5),(3.1.7),(3.1.9)and(3.1.11), the third order approximate solution by OHAM for, p
=
1, is;˜
u
(
x,
C1,
C2,
C3) =
u0(
x) +
u1(
x,
C1) +
u2(
x,
C1,
C2) +
u3(
x,
C1,
C2,
C3).
(3.1.12) Following the procedure described in Section2on the domain between a=
0 and b=
1, using the residual,R
= ˜
u000(
x,
C1,
C2,
C3) −
25u˜
0(
x,
C1,
C2,
C3) +
1,
(3.1.13) the following values of Ci’s are obtained:C1
= −
0.
587734742,
C2= −
0.
140650653,
C3=
0.
141306736.
By considering these values our approximate solution becomes,˜
u
(
x) = −
0.
0121071+
0.
0986628x2−
0.
166667x3+
0.
205258x4−
0.
205679x5+
0.
160292x6−
0.
0982484x7+
0.
0393381x8−
0.
0087418x.
9 (3.1.14) Note that by taking good zero order problem (may be the initial guess) one can meet sufficiently good accuracy at lower order solution.In theTable 1we compare the exact solution(3.1.3), OHAM solution(3.1.14), numerical method solution based on Pade0 approximants [25] (Fig. 1).
Table 2
x Exact solution Solution OHAM Error
0.0 0.00 0.00 0.00 0.1 0.0001 0.0001 −4.61510 E−18 0.2 0.0016 0.0016 −7.99457 E−18 0.3 0.0081 0.0081 −1.01810 E−17 0.4 0.0256 0.0256 −1.12175 E−17 0.5 0.0625 0.0625 −1.11477 E−17 0.6 0.1296 0.1296 −1.00163 E−17 0.7 0.2401 0.2401 −7.86921 E−18 0.8 0.4096 0.4096 −4.75372 E−18 0.9 0.6561 0.6561 −7.19025 E−19 1.0 1.00 1.00 4.18358 E−18
Error=Exact−OHAM.
3.2. Example 2
We consider in this example the fourth order nonlinear problem
u(4)
(
x) +
u(
x)
u0(
x) −
4x7−
24=
0,
with boundary conditions given at four different points,
u
(
0) =
0,
u000(
0.
25) =
6,
u00(
0.
5) =
3,
u(
1) =
1.
(3.2.1) The exact solution of this problem is,u
(
x) =
x4.
(3.2.2)Applying OHAM, we have the following zero, first and second order solutions:
u0
(
x) = −
0.
000451433x−
0.
0000523461x2−
1.
27157×
10−6x3+
x4+
O(
x8),
(3.2.3) u(
x,
C1) =
0.
000451433x+
8.
69509×
10−7C1x+
0.
0000523461x2+
5.
0327×
10−7C1x2+
1.
27157×
10−6x3+
7.
44941×
10−8C1x3+
1.
69826×
10−9C1x5+
O(
x8)
and (3.2.4) u2(
x,
C1,
C2) = −
0.
000450563C1x−
1.
06141×
10−10C12x+
8.
695089×
10 −7C 2x−
0.
0000518434C1x2−
5.
30707×
10−10C12x2+
5.
0327×
10−7C2x2−
1.
19813×
10−6C1x3−
1.
06141×
10−9C12x 3+
7.
44941×
10−8C2x3+
3.
39653×
10−9C1x5+
1.
69826×
10−9C12x 5+
1.
69826×
10−9C 2x5+
1.
96923×
10−10C1x6+
1.
96923×
10−10C2x6+
9.
25755×
10−12C1x7+
9.
25755×
10−12C2x7+
O(
x8).
(3.2.5)Using Eqs.(3.2.3)–(3.2.5), we obtain the following second order approximate solution by OHAM,
˜
u
(
x,
C1,
C2) =
u0(
x) +
u1(
x,
C1) +
u2(
x,
C1,
C2).
For a
=
0 and b=
1, following the procedure for values of Ci’s in Section2we get: C1=
2.
94458×
10−13,
C2=
9.
9164×
10−11.
By considering these values our solution becomes,
˜
u
(
x) =
5.
24705×
10−17x−
6.
39007×
10−17x2+
7.
05627×
10−18x3+
x4+
1.
6990×
10−19x5+
1.
95857×
10−20x6+
9.
20741×
10−17x7.
(3.2.6) In theTable 2we present values of the exact solution(3.2.2), OHAM solution(3.2.6)and the error appear between them (Fig. 2).3.3. Example 3
Let us consider the three point second order nonlinear problem
u00
(
x) +
3 8u(
x) +
2 1089u 02(
x) +
1=
0,
0≤
x≤
1,
u(
0) =
0,
u 1 3=
u(
1).
(3.3.1)Fig. 2. The solid line (red) denotes the OHAM solution while the dotted line denotes the exact solution for the domain−100≤x≤100; the plot shows excellent agreement with the exact solution.
Table 3a
x OHAM solution HPM solution ADM solution
0.0 0.00 0.00 0.00 0.1 0.0656 0.0656 0.0656 0.2 0.121 0.1209 0.1209 0.3 0.1659 0.1658 0.1658 0.4 0.2002 0.2001 0.2001 0.5 0.2237 0.2236 0.2236 0.6 0.2364 0.2363 0.2363 0.7 0.2382 0.2381 0.2382 0.8 0.2291 0.2291 0.2291 0.9 0.2092 0.2091 0.2091
The approximate second order OHAM solution is,
˜
u(
x,
C1,
C2) =
2 3−
6566C1 88209−
30015838C2 1 864536409−
3283C2 88209 x−
1 2−
8C1 9801−
104480C2 1 288178803−
4C2 9801 x2+
3235C1 39204+
179618441C12 4610860848+
3235C2 78408 x3−
3235C1 104544+
31628851C12 2049271488+
3235C2 209088 x4+
2030933C 2 1 2732361984x 5−
2030933C12 10929447936x 6.
(3.3.2)For a
=
0 and b=
1, the following values of Ci’s are obtained: C1=
1.
034937306,
C2= −
4.
141101662.
The approximate solution now becomes;
˜
u
(
x) =
0.
706567121x−
0.
500456984x2−
0.
043730752x3+
0.
01551444x4+
0.
000796133x5−
0.
000199033x6.
(3.3.3)In order to compare the solution of this problem, we solve the same problem by HPM. The HPM solution is;
u
˜
(
x) =
610695134 864536409x−
288440291 576357606x 2−
200856379 4610860848x 3+
31783619 2049271488x 4+
2030933 2732361984x 5−
2030933 10929447936x 6.
(3.3.4)Table 3a. displays values of the OHAM solution(3.3.3), HPM solution(3.3.4)and solution by ADM [2].
In theTable 3b. the residual R
= ˜
u(4)(
x) + ˜
u(
x)˜
u0(
x) −
4x7−
24 for OHAM solution and HPM solution is calculated for different values of x.The proposed method shows greater accuracy than HPM and ADM (Fig. 3).4. Conclusions
In this paper we have used OHAM to find the approximate analytic solution to multipoint boundary value problems. It is observed that the method is explicit, effective and reliable. It works well for both linear and nonlinear problems and represents the fastest convergence as well as a remarkable low error. The OHAM also provides us with very simple way to
Table 3b x R (OHAM) R (HPM) 0.0 2.904 E−6 9.018 E−6 0.1 1.103 E−6 1.004 E−4 0.2 −7.372 E−8 1.860 E−4 0.3 −5.032 E−7 2.612 E−4 0.4 −3.761 E−7 3.222 E−4 0.5 −9.272 E−9 3.659 E−4 0.6 2.949 E−7 3.901 E−4 0.7 3.448 E−7 3.936 E−4 0.8 1.115 E−7 3.762 E−4 0.9 −2.641 E−7 3.388 E−4 -4 -2 2 4 -5 -4 -3 -2 -1
Fig. 3. The solid line (red) denotes the OHAM solution while the dotted line denotes the HPM solution for the domain−4<x<5. control and adjust the convergence of the series solution using the auxiliary constants Ci’s which are optimally determined. The solution curve is very smooth and is amenable for any investigation and interpretation. Furthermore, the results of the method show excellent agreement with the exact solution. This method has a great potential to attract researchers, scientists and engineer of every field.
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